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Updated on: 07 Apr 2015, 04:40
Originally posted by Harley1980 on 06 Apr 2015, 11:19.
Last edited by Harley1980 on 07 Apr 2015, 04:40, edited 3 times in total.
Last edited by Harley1980 on 07 Apr 2015, 04:40, edited 3 times in total.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
22% (01:38) correct
78%
(01:29)
wrong
based on 192
sessions
History
Date
Time
Result
Not Attempted Yet
Is \((x+y)(x-y) =\) even integer?
1) \(x^2+y^2 =\) even
2) \(x+y =\)even
Source: self-made
1) \(x^2+y^2 =\) even
2) \(x+y =\)even
Source: self-made
07 Apr 2015, 02:08
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Harley1980
So now the question is - what is the actual answer?
Try using both statements together: If you know that \(x + y\) is an even integer and \(x^2 + y^2\) is an even integer, is it necessary that \(x^2 - y^2\) should be an even integer too? Actually, no!
It is easy to show that if x and y are both even/both odd integers, then \(x+y\) is even, \(x^2 + y^2\) is even and \(x^2 - y^2\) is even.
But how do you figure out a case where \(x+y\) is even, \(x^2 + y^2\) is even but \(x^2 - y^2\) is not an even integer?
Let me give you an example: Say \(x = 3+\sqrt{6}\) and \(y = 3 - \sqrt{6}\).
Here, \(x + y = 6\) (even integer)
\(x^2 + y^2 = 30\) (even integer)
But \(x^2 - y^2 = 12\sqrt{6}\)
So answer is (E).
What you should think about is: How can you prove with variables that answer is (E)? Perhaps, the form of x and y that I have given as an example can help you!
General Discussion
06 Apr 2015, 14:22
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Am i missing something simple?
A)(x+y)(x-y) = x^2+y^2 and we are told that is even... so suff
B) we are told (x+y) is even. even * even = even and odd * even = even . Even if (x-y) = 0, gmat considers 0 as even. So Suff
A)(x+y)(x-y) = x^2+y^2 and we are told that is even... so suff
B) we are told (x+y) is even. even * even = even and odd * even = even . Even if (x-y) = 0, gmat considers 0 as even. So Suff
It is better to make these errors during practice so that you do not make them during the actual exam!! 
